Problem 29 - Calculus Early Transcendentals

Question

Use linear approximations to estimate the following quantities. Choose a value of a that produces a small error. \(1 / \sqrt[3]{510}\)

Step-by-Step Solution

Answer

The approximate linear approximation for finding \(\frac{1}{\sqrt[3]{510}}\) is \(\frac{1}{8}+\frac{1}{3}\cdot 2^{-11}\).

Step 1: Find the function and its derivative

We need to find the derivative of the function \(f(x) = \frac{1}{\sqrt[3]{x}}\). First, rewrite the function as \(f(x) = x^{-1/3}\) to simplify the differentiation process. Now find the derivative: $$ f'(x) = -\frac{1}{3}x^{-4/3}. $$

Step 2: Choose an appropriate value for a

We are asked to approximate \(\frac{1}{\sqrt[3]{510}}\). We must choose a value for \(a\) close to \(510\) for which we can easily find \(f(a)\). A close and easy-to-work-with value is \(a=512\), since it's a power of 2. Specifically, \(512 = 2^9\) and \(\sqrt[3]{512} = 2^3 =8\). So let's use \(a=512\).

Step 3: Find \(f(a)\) and \(f'(a)\)

Now, we need to find \(f(512)\) and \(f'(512)\). We have: $$ f(512) = \frac{1}{\sqrt[3]{512}} = \frac{1}{8}, $$ and $$ f'(512) = -\frac{1}{3} (512)^{-4/3} = -\frac{1}{3}(2^9)^{-4/3} = -\frac{1}{3}(2^{-12}) = -\frac{1}{3}\cdot 2^{-12}. $$

Step 4: Compute the linear approximation

We have all the required values to compute the linear approximation: $$ L(x) = f(a) + f'(a)(x-a) = \frac{1}{8} - \frac{1}{3}\cdot 2^{-12} (x - 512). $$ Now substitute \(x = 510\) to find our approximation: $$ L(510) = \frac{1}{8} - \frac{1}{3}\cdot 2^{-12} (510 - 512) = \frac{1}{8} + \frac{1}{3}\cdot 2^{-12} \cdot 2 = \frac{1}{8}+\frac{1}{3}\cdot 2^{-11}. $$ This gives us: $$\frac{1}{\sqrt[3]{510}} \approx \frac{1}{8}+\frac{1}{3}\cdot 2^{-11}.$$

Key Concepts

Derivative Calculation

When we want to find the slope of a function at any given point, we must calculate its derivative. The exercise involves the function \(f(x) = \frac{1}{\sqrt[3]{x}}\), which can be rewritten as \(x^{-1/3}\) to make differentiation easier. This step transforms the problem into a more familiar form that follows basic power rule differentiation. The derivative of this function becomes:

• \(f'(x) = -\frac{1}{3}x^{-4/3}\).

The derivative provides us with the rate of change of the function with respect to \(x\). This information is crucial in linear approximations because it helps us understand the behavior of the function near the point of interest. In this example, knowing \(f'(x)\) allows us to estimate how the value of \(\frac{1}{\sqrt[3]{510}}\) changes as \(x\) shifts slightly from an easy-to-calculate point.

Cubic Root Estimation

Estimating cubic roots involves finding values that are close, yet manageable enough to compute without complex calculations. In this practice problem, we needed to estimate \(\frac{1}{\sqrt[3]{510}}\). We chose a nearby number, 512, which makes calculations smoother because its cubic root is an integer. Specifically, 512 equals \(2^9\), and its cube root is \(2^3 = 8\).
The benefit of choosing nearby estimates, like 512 here, is that it's easier to work with integers and common powers, reducing inevitable calculation errors in real-world scenarios. Such estimations are extremely useful when precise instruments or computers are not accessible, enabling quick mental calculations that are sufficiently accurate for everyday purposes.

Function Approximation

Function approximation allows us to predict the behavior of complex functions around a specific point. Linear approximation is a common technique, using tangents of the function to make our predictions. By applying the formula:

• \(L(x) = f(a) + f'(a)(x-a)\).

We substitute known values from our derivative calculation. Here, \(L(x)\) becomes the line that estimates \(f(x)\) near \(a\), where \(a\) is the value chosen near the point of interest.
In this case, linear approximation calculated \(L(510)\) by making use of the chosen \(a = 512\), considering both \(f(a) = \frac{1}{8}\) and \(f'(512)\), resulted in a very manageable approximation. These calculated approximations depict how a function behaves in small interval changes. This helps in quickly estimating complex calculations that would otherwise need higher computational resources.